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Calculation of the length and alternative form of the parameteric representation of a curve, involving trigonometric functions.

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You are given the curve $\\mathcal C$ be the space curve parametrically given by $\\boldsymbol{r} = \\pmatrix{\\var{a}\\cos(\\simplify{{b}t}),\\var{-a}\\sin(\\simplify{{b}t})}$, where $t_1\\pi\\leqslant t\\leqslant t_2\\pi$.

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a) The tangent vector to the curve is given by $\\boldsymbol{u} = \\frac{\\mathrm{d}r}{\\mathrm{d}t} \\equiv\\pmatrix{\\frac{\\mathrm{d}x}{\\mathrm{d}t},\\frac{\\mathrm{d}y}{\\mathrm{d}t}} =
\\pmatrix{\\simplify{{-a*b}sin({b}*t)}, \\simplify{{-a*b}*cos({b}*t)}} $ (note that, since the curve is not parametrised by arc length, $\\frac{\\mathrm{d}r}{\\mathrm{d}t}$ does not give the unit tangent!).

b) The length $s$ of the curve in the range $t_1\\pi\\leqslant t\\leqslant t_2\\pi$ is given by

\\[s = \\int_{t_1\\pi}^{t_2\\pi}{u\\mathrm{d}t},\\]

where $u^2=\\lvert\\boldsymbol{u}\\rvert^2=\\boldsymbol{u\\cdot u}$.

In this question, therefore, $\\boldsymbol{u}=\\pmatrix{\\frac{\\mathrm{d}}{\\mathrm{d}t}\\left\\{\\var{a}\\cos(\\simplify{{b}*t})\\right\\},\\frac{\\mathrm{d}}{\\mathrm{d}t}\\left\\{\\var{-a}\\sin(\\simplify{{b}*t})\\right\\}}=\\pmatrix{\\var{-a*b}\\sin(\\simplify{{b}*t}),\\var{-a*b}\\cos(\\simplify{{b}*t})}$, and so $u^2=\\var{(a*b)^2}$.

Then

\\[s=\\int_{t_1\\pi}^{t_2\\pi}{u\\mathrm{d}t}=\\var{a*b}\\int_{t_1\\pi}^{t_2\\pi}{\\mathrm{d}t}=\\var{a*b}(t_2-t_1)\\pi.\\]

Finally, substitute $t_1=-\\simplify{1/{t2}}$ and $t_2=\\simplify{1/{t2}}$ into this expression for $s$, to find the length of the curve over the given range of $t$.

Hence $s=\\simplify{{2*a*b}/{t2}}\\pi$.

c) An alternative parametric representation, using $s$ as the curve parameter is given by

\\[s=\\int_{t_1\\pi}^{t}{u\\mathrm{d}\\tau}=\\var{a*b}\\int_{t_1\\pi}^{t}{\\mathrm{d}\\tau}=\\var{a*b}(t-t_1\\pi).\\]

Now rearrange this expression for $t(s)$, so

\\[t(s)=\\frac{s}{\\var{a*b}}+t_1\\pi,\\]

and substitute into the original representation of the curve $\\mathcal C\\equiv\\pmatrix{\\var{a}\\cos(\\simplify{{b}t}),\\var{-a}\\sin(\\simplify{{b}t})}$ with $t_1\\pi\\leqslant t\\leqslant t_2\\pi$. Hence

\\[\\mathcal C \\equiv\\pmatrix{\\var{a}\\cos\\left(\\simplify{s/{a}}+\\var{b}t_1\\pi\\right),\\var{-a}\\sin\\left(\\simplify{s/{a}}+\\var{b}t_1\\pi\\right)},\\]

with $0\\leqslant s\\leqslant\\var{a*b}(t_2-t_1)\\pi$.

Finally, substitute $t_1=-\\simplify{1/{t2}}$ and $t_2=\\simplify{1/{t2}}$ into the above expressions, to find the specific parametric representation corresponding to the given range of t:

\\[\\mathcal C \\equiv\\pmatrix{\\var{a}\\cos\\left(\\simplify{s/{a}}-\\simplify{{b*pi}/{t2}}\\right),\\var{-a}\\sin\\left(\\simplify{s/{a}}-\\simplify{{b*pi}/{t2}}\\right)},\\]

with $0\\leqslant s\\leqslant \\simplify{{2*a*b}/{t2}}\\pi$.

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Find the tangent vector $\\boldsymbol{u}$ to the curve.

$\\boldsymbol{u}=($[[0]]$,$[[1]]$)$. (Do not enter decimals in your answers.)

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Find the length $s$ of the curve between $t_1=-\\simplify{1/{t2}}$ and $t_2=\\simplify{1/{t2}}$.

$s=$ [[0]]. (Enter your answer as a fractional multiple of $\\pi$. Do not enter decimals.)

Do not enter decimals in your answer.

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Find another parametric representation of the curve, again with $t_1=-\\simplify{1/{t2}}$ and $t_2=\\simplify{1/{t2}}$, using $s$ as the curve parameter, such that $0\\leqslant s\\leqslant \\simplify{{2*a*b}/{t2}}\\pi$.

$s\\rightarrow($[[0]]$,$[[1]]$)$. (Enter your answers as fractional multiples of $\\pi$. Do not enter decimals.)

Enter your answer as a fractional multiple of $\\pi$. Do not enter decimals.

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Enter your answer as a fractional multiple of $\\pi$. Do not enter decimals.

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